Question: A function $f:\mathbb{Z} \to \mathbb{Z}$ satisfies
\begin{align*}
f(x+4)-f(x) &= 8x+20, \\
f(x^2-1) &= (f(x)-x)^2+x^2-2
\end{align*}for all integers $x.$  Enter the ordered pair $(f(0),f(1)).$
Setting $x = 0$ in the second equation, we get
\[f(-1) = f(0)^2 - 2.\]Setting $x = -1$ in the second equation, we get
\[f(0) = (f(-1) + 1)^2 - 1.\]Let $a = f(0)$ and $b = f(-1)$; then $b = a^2 - 2$ and $a = (b + 1)^2 - 1.$  Substituting $b = a^2 - 2,$ we get
\[a = (a^2 - 1)^2 - 1.\]This simplifies to $a^4 - 2a^2 - a = 0,$ which factors as $a(a + 1)(a^2 - a - 1) = 0.$  The quadratic $a^2 - a - 1 = 0$ has no integer solutions, so $a = 0$ or $a = -1.$

Suppose $f(0) = a = 0.$  Then $f(-1) = -2.$  Setting $x = -1$ in the first equation, we get
\[f(3) - f(-1) = 12,\]so $f(3) = f(-1) + 12 = 10.$  But setting $x = 2$ in the second equation, we get
\[f(3) = (f(2) - 2)^2 + 2,\]so $(f(2) - 2)^2 = 8.$  No integer value for $f(2)$ satisfies this equation.

Therefore, $f(0) = a = -1.$  Setting $x = 1$ in the second equation, we get
\[f(0) = (f(1) - 1)^2 - 1,\]so $(f(1) - 1)^2 = 0,$ which forces $f(1) = 1.$

Hence, $(f(0),f(1)) = \boxed{(-1,1)}.$  Note that the function $f(n) = n^2 + n - 1$ satisfies the given conditions.